(a) A cantilever beam ABC as shown in the above figure is having a total span of 2·0 m. The maximum safe allowable bending stress…

(a) The cantilever is fixed at A. Segment AB (1.0 m) has section 1-1, 0.20 m wide × 0.45 m deep; segment BC (1.0 m) has section 2-2, 0.20 m × 0.20 m. With a UDL w (kN/m) over the full 2 m, the bending moment at distance x from the free end C is wx²/2. Because the section changes, the maximum stress need not occur at the support, so both the support A and the step B must be checked.

Section moduli (Z = bd²/6):

• Z₁ = 0.20 × 0.45²/6 = 6.75 × 10⁻³ m³ (AB)
• Z₂ = 0.20 × 0.20²/6 = 1.333 × 10⁻³ m³ (BC)

Check at the support A: M_A = w × 2²/2 = 2w kN·m, so σ_A = 2w/6.75 × 10⁻³ = 296.3w kN/m². Setting σ_A ≤ 7500 gives w ≤ 25.3 kN/m.

Check at the step B (just inside the smaller section): M_B = w × 1²/2 = 0.5w kN·m, so σ_B = 0.5w/1.333 × 10⁻³ = 375w kN/m². Setting σ_B ≤ 7500 gives w ≤ 20 kN/m.

The smaller section governs: safe UDL w = 20 kN/m. At this load the stress at A is only 2 × 20/6.75 × 10⁻³ = 5926 kN/m², while just right of B it reaches the full 7500 kN/m².

Shear at A: V_A = 20 × 2 = 40 kN. For a rectangle the shear stress is parabolic across the depth, with its maximum at the neutral axis:

τ_max = 1.5V/(bd) = 1.5 × 40/(0.20 × 0.45) = 666.7 kN/m² ≈ 0.67 N/mm² (the average is 444.4 kN/m²).

(b) Notation: b = width, d = effective depth, x_u = depth of the neutral axis, A_st = area of tension steel, f_ck = characteristic cube strength, f_y = characteristic yield stress of steel.

Assumptions (IS 456, cl. 38.1):

• Plane sections remain plane, so strain varies linearly with depth.
• The maximum compressive strain in concrete is ε_cu = 0.0035.
• Concrete in tension is ignored.
• Concrete follows Fig. 1 with design peak stress 0.67f_ck/γ_m = 0.67f_ck/1.5 = 0.446f_ck: parabolic up to a strain of 0.002, then constant to 0.0035.
• Steel follows Fig. 2 with design yield stress f_y/γ_s = f_y/1.15 = 0.87f_y.

Concrete compression. Since the strain is linear, 0.002 occurs at depth x_u(1 − 0.002/0.0035) = 3x_u/7 below the top. The stress block therefore has a rectangle of height 3x_u/7 and a parabola of height 4x_u/7 (area = ⅔ × height × peak stress):

C = 0.446f_ck b [3x_u/7 + ⅔ × 4x_u/7] = 0.446f_ck b x_u (17/21) = 0.36 f_ck b x_u.

Taking moments of the two parts about the top fibre gives the line of action of C at 0.42 x_u from the compression face, so the lever arm is z = d − 0.42x_u.

Steel tension and strain compatibility. From the strain diagram, ε_s = 0.0035(d − x_u)/x_u. For the steel to reach its design stress 0.87f_y, IS 456 requires ε_s ≥ 0.87f_y/E_s + 0.002 (the 0.002 allows for the gradual yield of cold-worked bars in Fig. 2). Then T = 0.87 f_y A_st.

Equilibrium C = T:

0.36 f_ck b x_u = 0.87 f_y A_st, so x_u = 0.87 f_y A_st / (0.36 f_ck b), or x_u/d = 0.87 f_y A_st / (0.36 f_ck b d).

Limit on x_u: setting ε_s = 0.87f_y/E_s + 0.002 in the compatibility equation gives x_u,max/d = 0.0035/(0.0055 + 0.87f_y/E_s) = 0.53, 0.48 and 0.46 for Fe 250, Fe 415 and Fe 500. If the computed x_u exceeds x_u,max the section is over-reinforced and the steel does not yield.

Moment of resistance: M_u = 0.87 f_y A_st (d − 0.42 x_u) = 0.36 f_ck b x_u (d − 0.42 x_u).

(c) Each angle carries P = 540/2 = 270 kN along its C.G., which is 29.2 mm from the heel edge (weld l₁) and 100 − 29.2 = 70.8 mm from the toe edge (weld l₂).

Balancing the welds. The two weld forces F₁ and F₂ must have their resultant on the line of P:

• F₁ + F₂ = 270 and F₁ × 29.2 = F₂ × 70.8.
• F₁ = 270 × 70.8/100 = 191.2 kN (heel weld, nearer the C.G.).
• F₂ = 270 × 29.2/100 = 78.8 kN (toe weld).

Weld strength (IS 800, cl. 10.5.7):

• f_wd = f_u/(√3 γ_mw) = 410/(√3 × 1.25) = 189.4 N/mm².
• Throat t_t = 0.7 × 6 = 4.2 mm.
• Strength per mm = 4.2 × 189.4 = 795 N/mm = 0.795 kN/mm.

Lengths:

• Effective l₁ = 191.2/0.795 = 240.4 mm; effective l₂ = 78.8/0.795 = 99.1 mm.
• Overall length = effective length + 2s (cl. 10.5.4): l₁ = 240.4 + 12 = 252.4 mm and l₂ = 99.1 + 12 = 111.1 mm.
• Provide l₁ = 255 mm and l₂ = 115 mm for each angle.

Checks: both lengths exceed 4s = 24 mm and 40 mm; the 6 mm weld is above the 5 mm minimum for a 12 mm part and below the ¾ × 12 = 9 mm limit at the rounded toe; the welds are shorter than 150t_t = 630 mm, so no long-joint reduction is needed.